Oscillations Test

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Oscillations Test - 57

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Question 1
1 / -0

Two masses $$m_1$$ and $$m_2$$ are suspended together by a massless springs of constant $$k$$. When the masses are in equilibrium, $$m_1$$ is removed without disturbing the system. Then the angular frequency of oscillation of $$m_2$$ is

A
$$\sqrt{\dfrac{k}{m_1}}$$

B
$$\sqrt{\dfrac{k}{m_2}}$$

C
$$\sqrt{\dfrac{k}{m_1+m_2}}$$

D
$$\sqrt{\dfrac{k}{m_1m_2}}$$

Solution

When only the mass $$m_2$$ is suspended let the elongation of the spring be $$x_1$$. When both the masses $$(m_2 + m_1)$$ together are suspended, the elongation of the spring is $$(x_1 + x_2)$$. Thus, we have
$$m_2g=kx_1$$
and, $$(m_1+m_2)g=k(x_1+x_2)$$, where k is the spring constant.
$$\Rightarrow m_1g+m_2g =kx_1 + kx_2$$
$$\Rightarrow m_1=kx_2$$, since $$m_2g=kx_1$$.

Thus, $$x_2$$ is the elongation of the spring due to the mass $$m_1$$ only. When the mass $$m_1$$ is removed the mass $$m_2$$ executes SHM with the amplitude $$x_2$$.
Amplitude of vibration, $$x_2 = m_1g$$
Hence, angular frequency, $$\omega = \sqrt{\frac{k}{m_2}}$$ is our required answer.
Question 2
1 / -0

What is constant in S.H.M.

A
Restoring force

B
Kinetic energy

C
Potential energy

D
Periodic time

Solution

The only thing that remains constant for one particle performing SHM is its periodic time or simply time period.
Question 3
1 / -0

A mass $$m$$ is suspedend from the two coupled springs connected in series. The force constant for springs are $$K_1$$ and $$K_2$$. The time period of the suspended mass will be

A
$$T=2\pi \sqrt{\left(\dfrac{m}{K_1+K_2}\right)}$$

B
$$T=2\pi \sqrt{\left(\dfrac{m}{K_1+K_2}\right)}$$

C
$$T=2\pi \sqrt{\left(\dfrac{m(K_1+K_2)}{K_1K_2}\right)}$$

D
$$T=2\pi \sqrt{\left(\dfrac{mK_1K_2}{K_1+K_2}\right)}$$

Solution

In Series $$K_{eq}=\dfrac{K_1K_2}{K_1+K_2}$$
so Time period $$T=\sqrt{\dfrac{m}{K_{eq}}}=2\pi \sqrt{\dfrac{m(K_1+K_2)}{K_1K_2}}$$
Question 4
1 / -0

A system exhibiting $$S.H.M$$ must possess

A
Inertia only

B
Elasticity as well as inertia

C
Elasticity inertia and an external force

D
Elasticity only

Solution

Basic conditions to execute simple harmonic motion are: (i) There must be an elastic restoring force acting on the system, (ii) the system must have inertia, and (iii) the acceleration of the system should be directly proportional to its displacement and is always directed to mean position.
Question 5
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Which of the following is not true? In the case of a simple pendulum for small amplitudes the periodic of oscillation is

A
Directly proportional to square root of the length of the pendulum

B
Inversely proportional to square root of the acceleration due to gravity

C
Dependent on the mass, size and material of the bob

D
Independent of the amplitude

Solution

Time period is given by, $$T=2\sqrt{\frac{l}{g}}$$
Clearly, $$T\propto \sqrt{l}$$. So, option A is correct.
Also, $$T\propto\frac{1}{\sqrt{g}}$$ So, option B is also correct.
We don't see any amplitude term in the time period equation hence, its independent of amplitude. So, option D is also correct.
We don't see any mass or size or material dependency in the time period equation. So, its not dependent upon mass or size or material of the bob. Hence, option C is the wrong statement here.
Question 6
1 / -0

A simple pendulum is sec up in a trolley which moves to the right with an acceleration $$a$$ on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle $$\theta$$ with the vertical

A
$$\tan^{-1} \dfrac ag$$ in the forward direction

B
$$\tan^{-1} \dfrac ag$$ in the backward direction

C
$$\tan^{-1} \dfrac ga$$ in the backrward direction

D
$$\tan^{-1} \dfrac ga$$ in the forward direction

Solution

In acceleration frame of reference a frictions force (pseudo force) $$ma$$ acts on the bob of pendulum as shown in figure
Hence,
$$\tan \theta =\dfrac {ma}{mg}=\dfrac ag$$
$$\Rightarrow $$
$$\theta =\tan^{-1}\left(\dfrac ag \right)$$ in the backward direction.
Question 7
1 / -0

A particle moving along the $$x-$$axis execute simple harmonic motion, then the force acting on it is given by

A
$$-A\ Kx$$

B
$$A\ \cos \ (Kx)$$

C
$$A\ exp \ (-Kx)$$

D
$$A\ Kx$$

Solution

For SHM, we need $$F\propto -x$$, which means that force will act in the opposite direction to force.
Looking at the given options we see that only $$F=-AKx$$ will be satisfying the condition of SHM if we assume $$AK$$ to be the constant of proportionality.
Question 8
1 / -0

The period of simple pendulum is measured as $$T$$ in a stationary lift. If the lift moves upward with an acceleration of $$5\ g$$, the period will be

A
The same

B
Increased by $$3/5$$

C
Decreased by $$2/3$$ times

D
None of the above

Solution

$$T=2\pi\sqrt{\dfrac{l}{g}}\\T\propto \sqrt{\dfrac{1}{g}}$$
$$\dfrac {T'}{T}=\sqrt {\dfrac {g}{g+a}}=\sqrt {\dfrac {g}{g+5g}}=\sqrt {\dfrac {1}{6}}$$

$$\Rightarrow T' =\dfrac {T}{\sqrt 6}$$
Question 9
1 / -0

Mark the wrong statement

A
All S.H.M.s have fixed time period

B
All motion having same time period are S.H.M.

C
In S.H.M. total energy is proportional to square of amplitude

D
Phase constant of S.H.M. depends upon initial conditions

Solution

For SHM, the force of the particle is given by,
$$F=-kx$$
$$\Rightarrow ma=-kx$$
$$\Rightarrow a=\frac{-k}{m} x$$
We also have,
$$ a=-\omega^2 x$$
From this, we can say $$\omega^2=\dfrac{k}{m}$$
or, $$\omega^2=\sqrt{\dfrac{k}{m}}$$, which is a constant.
Now, time period is given by, $$T=\dfrac{2\pi}{\omega}$$ which will also be a constant. Hence, option A is right.

All motion which have same period may not be SHM. A good example is that of when a particle is moving in a circle. This case has same time period but is not an example of SHM. Hence, the wrong answer is B.

The total energy of SHM is given by, $$KE=\dfrac{1}{2}kA^2$$. So, we can see that the total energy is directly proportional to the square of amplitude. Hence, option C is also correct.

The phase constant is given by $$y=Asin\omega t + \phi$$ where $$\phi$$ is the phase constant. Clearly we can see that the phase constant depends directly upon the initial condition of the particle. Hence, option D is also correct.
Question 10
1 / -0

The time period of a simple pendulum is $$2$$ sec. If its length is increased $$4$$ times, then its period becomes

A
$$16$$ sec

B
$$12$$ sec

C
$$8$$ sec

D
$$4$$ sec

Solution

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Re-Attempt Weekly Quiz Competition

Oscillations Test - 57

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Oscillations Test - 57

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SHARING IS CARING

Question 1

$$\sqrt{\dfrac{k}{m_1}}$$

$$\sqrt{\dfrac{k}{m_2}}$$

$$\sqrt{\dfrac{k}{m_1+m_2}}$$

$$\sqrt{\dfrac{k}{m_1m_2}}$$

Solution

Question 2

Restoring force

Kinetic energy

Potential energy

Periodic time

Solution

Question 3

$$T=2\pi \sqrt{\left(\dfrac{m}{K_1+K_2}\right)}$$

$$T=2\pi \sqrt{\left(\dfrac{m}{K_1+K_2}\right)}$$

$$T=2\pi \sqrt{\left(\dfrac{m(K_1+K_2)}{K_1K_2}\right)}$$

$$T=2\pi \sqrt{\left(\dfrac{mK_1K_2}{K_1+K_2}\right)}$$

Solution

Question 4

Inertia only

Elasticity as well as inertia

Elasticity inertia and an external force

Elasticity only

Solution

Question 5

Directly proportional to square root of the length of the pendulum

Inversely proportional to square root of the acceleration due to gravity

Dependent on the mass, size and material of the bob

Independent of the amplitude

Solution

Question 6

$$\tan^{-1} \dfrac ag$$ in the forward direction

$$\tan^{-1} \dfrac ag$$ in the backward direction

$$\tan^{-1} \dfrac ga$$ in the backrward direction

$$\tan^{-1} \dfrac ga$$ in the forward direction

Solution

Question 7

$$-A\ Kx$$

$$A\ \cos \ (Kx)$$

$$A\ exp \ (-Kx)$$

$$A\ Kx$$

Solution

Question 8

The same

Increased by $$3/5$$

Decreased by $$2/3$$ times

None of the above

Solution

Question 9

All S.H.M.s have fixed time period

All motion having same time period are S.H.M.

In S.H.M. total energy is proportional to square of amplitude

Phase constant of S.H.M. depends upon initial conditions

Solution

Question 10

$$16$$ sec

$$12$$ sec

$$8$$ sec

$$4$$ sec

Solution

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